\lecture{16}{2023-12-08}{}

$X$ is always compact metrizable.

\begin{theorem}
    Every minimal isometric flow $(X,\Z)$
    for $X$ a compact metrizable space%
    \footnote{Such a flow is uniquely determined by $h\colon X \to  X, x \mapsto 1\cdot x$.}
    is isomorphic to an abelian group rotation
    $(K, \Z)$, with
    $K$ an abelian compact group
    and some fixed $\alpha \in K$ such
    that $h(x) = x + \alpha$ for all $x \in K$
\end{theorem}
% \begin{example}
%     Let $\alpha \in S^1$
%     and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
% \end{example}
\begin{proof}
\gist{%
    The action of $1$ determines $h$.
    Consider
    \[
        \{h^n : n \in \Z\}  \subseteq  \cC(X,X)\gist{ = \{f\colon  X \to X : f \text{ continuous}\}}{},
    \]
    where the topology is the uniform convergence topology.
    Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
    Since the family $\{h^n : n \in \Z\}$ is uniformly equicontinuous,
    i.e.~
    \[
    \forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon,
    % Here we use isometric
    \]
    we have by the Arzel\`a-Ascoli-Theorem % TODO REF
    that $G$ is compact.

    $G$ is a closure of a topological group,
    hence it is a topological group,
    by \yaref{fact:topsubgroupclosure}.
    Since $h^n$ and $h^m$ commute for all $n, m \in \Z$,
    we obtain that $G$ is abelian.

    Take any $x \in X$ and consider the orbit $G \cdot x$.
    Since $\Z \acts X$ is minimal,
    i.e.~every orbit is dense,
    we have that $G \cdot x$ is dense in $X$.

    \begin{claim}
        $G \cdot x$ is compact.
    \end{claim}
    \begin{subproof}
        Since $\Z \acts X$ is continuous,
        $g \mapsto g x$ is continuous:

        Let $g_n$ be a sequence in $G$
        such that $g_n \to g$.
        Then $g_n x \to  gx$,
        since the topology on $\cC(X,X)$
        is the uniform convergence topology.

        Therefore the compactness of $G$ implies
        that the orbit  $Gx$ is compact.
    \end{subproof}

    Since $G\cdot x$ is compact and dense,
    we get $G \cdot x = X$
    (compact subsets of Hausdorff spaces are closed).

    Let $G_x = \{f \in G : f(x) = x\} <  G$
    be the stabilizer subgroup.
    Note that $G_x \subseteq G$ is closed.
    Take $K \coloneqq \faktor{G}{G_x}$ with the quotient topology.

    There is a continuous bijection
    \begin{IEEEeqnarray*}{rCl}
       K &\longrightarrow & X \\
       f G_x &\longmapsto & f(x).
    \end{IEEEeqnarray*}
    By compactness this is a homeomorphism,
    so this is an isomorphism between flows.

    For $\alpha = h$ we get that
    a flow $\Z \acts X$ corresponds to $\Z \acts K$
    with $(1,x) \mapsto x + \alpha$.
}{
    \begin{itemize}
        \item $G \coloneqq  \overline{\{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
        \item $G$ is compact (Arzela-Ascoli), abelian topological group (closure of ab. top. group)
        \item Take any $x \in G$.
        \item $Gx$ is compact (since $g \mapsto gx$ is continuous and $G$ is compact)
        \item Stabilizer $G_x$ is closed. $K \coloneqq  \faktor{G}{G_x}$, $K \to X, fG_x \mapsto f(x)$.
    \end{itemize}
}
\end{proof}
\begin{definition}
    Let $(X,T)$ be a flow
    and  $(Y,T)$ a factor of  $(X,T)$.
    Suppose there is $\eta \in \Ord$
    such that for any $\xi < \eta$
    there is a factor $(X_\xi, T)$ of  $(X,T)$
    with factor map $\pi_\xi\colon X \to  X_\xi$
    such that
    \begin{enumerate}[(a)]
        \item $(X_0, T) = (Y,T)$ and $(X_\eta, T) = (X,T)$.
        \item If  $\xi < \xi'$, then $(X_\xi, T)$ is a factor of  $(X_{\xi'}, T)$
            ``inside $(X,T)$'', i.e.~$\pi_\xi = \pi_{\xi, \xi'} \circ \pi_{\xi'}$,
            where $\pi_{\xi,\xi'}\colon X_{\xi'} \to  X_\xi$
            is the factor map.
        \item $\forall \xi <  \eta.~ (X_{\xi + 1}, T)$ is an isometric extension of $(X_\xi, T)$.
        \item $\xi \le \eta$ is a limit, then $(X_\xi, T)$
            is a limit of  $\{(X_\alpha,T), \alpha < \xi\}$.
    \end{enumerate}
    % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzAsNSwiWSJdLFsxLDIsIlhfe1xceGknfSJdLFsxLDMsIlhfXFx4aSJdLFswLDEsIlxccGkiLDAseyJjdXJ2ZSI6NH1dLFswLDIsIlxccGlfe1xceGknfSIsMix7ImN1cnZlIjotMX1dLFswLDMsIlxccGlfe1xceGl9IiwyLHsiY3VydmUiOjJ9XSxbMiwzLCJcXHBpX3tcXHhpLCBcXHhpJ30iXV0=
\[\begin{tikzcd}
	X \\
	\\
	& {X_{\xi'}} \\
	& {X_\xi} \\
	\\
	Y
	\arrow["\pi", curve={height=24pt}, from=1-1, to=6-1]
	\arrow["{\pi_{\xi'}}"', curve={height=-6pt}, from=1-1, to=3-2]
	\arrow["{\pi_{\xi}}"', curve={height=12pt}, from=1-1, to=4-2]
	\arrow["{\pi_{\xi, \xi'}}", from=3-2, to=4-2]
\end{tikzcd}\]

    Then we say that $(X,T)$ is a \vocab{quasi-isometric extension}
    of $(Y,T)$.
\end{definition}
\begin{definition}
    If $(Y,T)$ is trivial, i.e.~$|Y| = 1$,
    then a quasi-isometric extension $(X,T)$ of $(Y,T)$
    is called a \vocab{quasi-isometric flow}.
\end{definition}

\begin{corollary}
    Every quasi-isometric flow is distal.
\end{corollary}
\begin{proof}
    The trivial flow is distal.
    Apply \yaref{prop:isomextdistal} and \yaref{prop:limitdistal}.
\end{proof}

\begin{theorem}[Furstenberg]
    \label{thm:furstenberg}
    Every minimal distal flow is quasi-isometric.
\end{theorem}
By Zorn's lemma, this will follow from
\begin{theorem}[Furstenberg]
    \label{thm:l16:3}
    Let $(X, T)$ be a minimal distal flow
    and let $(Y, T)$ be a proper factor.%
    \footnote{i.e.~$(X,T)$ and $(Y,T)$ are not isomorphic}
    Then there is another factor $(Z,T)$ of $(X,T)$
    which is a proper isometric extension of  $Y$.

    % https://q.uiver.app/#q=WzAsMyxbMiwwLCIoWCxUKSJdLFswLDAsIihZLFQpIl0sWzEsMSwiKFosVCkiXSxbMSwwXSxbMiwwXSxbMSwyLCJcXHRleHR7aXNvbWV0cmljIGV4dGVuc2lvbn0iLDFdXQ==
\[\begin{tikzcd}
	{(Y,T)} && {(X,T)} \\
	& {(Z,T)}
	\arrow[from=1-1, to=1-3]
	\arrow[from=2-2, to=1-3]
\arrow["{\text{isometric extension}}"{description}, from=1-1, to=2-2]
\end{tikzcd}\]
\end{theorem}
\yaref{thm:furstenberg} allows us to talk about ranks of distal minimal flows:
\begin{definition}[{\cite[{}13.1]{Furstenberg}}]
    \label{def:floworder}
    Let $(X,T)$ be a quasi-isometric flow,
    and let $\eta$ be the smallest ordinal
    such that there exists a quasi-isometric system $\{(X_\xi, T), \xi \le \eta\}$
    with $(X,T) = (X_\xi, T)$.
    Then $\eta$ is called the \vocab{rank} or \vocab{order} of the flow
    and is denoted by $\rank((X,T))$.
\end{definition}

\begin{definition}+
    Let $X$ be a topological space.
    Let $K(X)$ denote the set of all compact subspaces of $X$
    and $K(X)^\ast \coloneqq  K(X)\setminus \{\emptyset\}$.
    If $d \le 1$ is a metric on $X$,
    we can equip $K(X)$ with a metric $d_H$ given by
    \begin{IEEEeqnarray*}{rClr}
        d_H(\emptyset, \emptyset) &\coloneqq & 0,\\
        d_H(K, \emptyset) &\coloneqq & 1 & K \neq \emptyset,\\
        d_H(K_0, K_1) &\coloneqq &
        \max \{\max_{x \in K_0}d(x,K_1), \max_{x \in K_1} d(x,K_0)\} &
        K_0,K_1 \neq \emptyset.
    \end{IEEEeqnarray*}

    The topology induced by the metric
    is given by basic open subsets\footnote{cf.~\yaref{s9e2}}
    of the form
    $[U_0; U_1,\ldots, U_n]$,
    for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
    where
    \[
        [U_0; U_1,\ldots,U_n] \coloneqq
        \{K \in K(X) | K \subseteq U_0 \land \forall 1\le i\le n.~K \cap U_i \neq \emptyset\}.
    \]
\end{definition}

We want to view flows as a metric space.
For a fixed compact metric space $X$,
we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$.
Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}}
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO

However we do not want to consider only flows on a fixed space $X$,
but we want to look all flows at the same time.
The Hilbert cube $\bH = [0,1]^{\N}$
embeds all compact metric spaces.
Thus we can consider $K(\bH)$,
the space of compact subsets of $\bH$.
$K(\bH)$ is a Polish space.\footnote{cf.~\yaref{s9e2}, \yaref{s12e4}}
% TODO LEARN EXERCISES
Consider $K(\bH^2)$.
A flow $\Z \acts X$ corresponds to the graph of
\begin{IEEEeqnarray*}{rCl}
    X &\longrightarrow & X \\
     x&\longmapsto & 1 \cdot x
\end{IEEEeqnarray*}
and this graph is an element of $K(\bH^2)$.

\begin{theorem}[Beleznay-Foreman]
    \label{thm:beleznay-foreman}
    Consider $\Z$-flows.
    \begin{itemize}
        \item For any $\alpha < \omega_1$,
            there is a distal minimal flow of rank $\alpha$.
        \item Distal flows form a $\Pi^1_1$-complete set,
            where flows are identified
            with their graphs as elements
            of $K(\bH^2)$ as above.
        \item Moreover, this rank is a $\Pi^1_1$-rank.
    \end{itemize}
\end{theorem}
